phpoperator-precedence

How does operator precedence affect the evaluation of $x = false && print 'printed' || true;?

I'm trying to understand why this expression:

$x = false && print 'printed' || true;

results in $x being false instead of true. I know that && has higher precedence than ||, but I'm confused about how the combination of &&, ||, and print works here. Can someone explain the exact evaluation process, step by step, in terms of operator precedence and short-circuit evaluation?

Solution

  • With an example from the PHP manual for print and the comments from @Jakkapong Rattananen, @shingo and @ Sergey Ligus I came to the following conclusion:

    As print is a language construct, evaluation is a bit different there. The example is the following:

    print "hello " && print "world";
// outputs "world1"; print "world" is evaluated first,
// then the expression "hello " && 1 is passed to the left-hand print

    This is similar to your expression, but differs in the way that two prints are used.

    So when the print after the && is evaluated first (and always returns 1), then probably the next operator || is evaluated next (left to right), so the whole expression is evaluated as if it were written

    $x = false && (print 'printed' || true);

    If you wrote it just a little bit different, it evaluates as you expected it, just according to operator precedence:

    $x = false && 1 || true;

    Here $x is false. So the difference is really the "language construct" nature of print.